Question: Graph this system of equations and solve. $4x+6y = 12$ $-8x+6y = -24$ 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 Click and drag the points to move the lines.
Convert the first equation, $4x+6y = 12$ , to slope-intercept form. $y = -\dfrac{2}{3} x + 2$ The y-intercept for the first equation is $2$ , so the first line must pass through the point $(0, 2)$ The slope for the first equation is $-\dfrac{2}{3}$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) You must also move $3$ positions to the right. $3$ positions to the right. $2$ positions down from $(0, 2)$ is $(3, 0)$ Graph the blue line so it passes through $(0, 2)$ and $(3, 0)$ Convert the second equation, $-8x+6y = -24$ , to slope-intercept form. $y = \dfrac{4}{3} x - 4$ The y-intercept for the second equation is $-4$ , so the second line must pass through the point $(0, -4)$ The slope for the second equation is $\dfrac{4}{3}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move up You must also move $3$ positions to the right. $3$ positions to the right. $4$ positions up from $(0, -4)$ is $(3, 0)$ Graph the green line so it passes through $(0, -4)$ and $(3, 0)$ The solution is the point where the two lines intersect. The lines intersect at $(3, 0)$.